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Help ... Geometry! My Bad!

I had recently opened a discussion thread about a geometry problem from an online-math-camp-test after the test was over. However, it turns out I broke some rule. There was a 24-hour rule ("no problems are to be discussed until 24 hours have passed") and I somehow missed that rule. Now that 24 hours have passed, it's completely okay to discuss about that problem.

I've deleted the previous discussion thread and I'm really sorry for the inconvenience.

Here's the problem I was talking about:

In \(△XYZ\),

\(2∠X=3∠Y\) and \(YZ=x\), \(XZ=y\), \(XY=z\). Find a triangle that has positive integer side lengths and satisfies the equation:

Thanks for the help. Notice that the angles don't have to be integers. So, trying to find cases when sine function has 'good looking' rational & irrational values obviously won't work. I tried using the cosine law like Timmy but the equation gets really messy. So I gave up.

I admit that angles don't need to be integers,but a conclusion can be made:

If the sines have ugly values,obviously no \(k\) will work for all 3 of them (by inspection), so we limit ourselves to rationals & those irrationals (for which we can easily define a conjugate, in the prospect of multiplying it to \(k\) to receive an integral value). Trial shows no such rational values exist (sine function has many less rational values, had it been tan function,it would have been less difficult!) & trying to find such irrational values which have similar conjugates for such \(\theta\) so that all angles are less than \(180^{\circ}\) also doesn't come up.

Here's a starter. Currently I have not been able to complete the solution, but I will (probably) post it soon. I am not sure if this is the most elegant way to do this though: it is certainly very tedious.

This is the way I approached the problem. But I gave up when I saw the ugly-looking equation. It looks like you're onto something. So post your solution solution when you're done. This problem has been nagging me for 3 months. It's time to put it to rest.

The only reason I think it might work is that it helps for the cancellation of \(y^2\). I expanded the L.H.S, and as expected, the \(z^2\) term also cancels out. But what remains is a mammoth equation, which is very difficult to interpret. So I'd say let's not expand the brackets, and make a few more trigonometric substitutions to see what we get. What I have got so far is not even close to a concrete solution.

I'm not sure if this would work: first solve the equation as a diophantine equation. there may be infinity number of solutions or only one may be. Narrow down the possible solutions by triangle inequality. Then check out for which of those solutions, 2X=3Y is true. [ i really am not good at math, so sorry if it is another silly method ]. By the way, I live in NAM garden, near to you.

I'm not spying on you. I first came to know about you and the fact that your college is Notre Dame from Brilliant. One day Tushar(you know him) was talking something about a boy named Mursalin who studies in Notre Dame at our English coaching. So I thought it might have been the same Mursalin from Brilliant and so I asked some things about you out of 'just' curiosity (not for spying) and came to know.... btw, was my suggestion able to help you out? Even a bit? Really sorry to waste your time if it didn't.